S OLVENT S ELECTION FOR S EPARATION Processes Requiring Solvents Extraction Partition Fractional Crystallization Extractive Distillation Liquid Chromatography

SOLVENT SELECTION FOR SEPARATION

Processes Requiring Solvents Extraction Partition Fractional Crystallization Extractive Distillation Liquid Chromatography Gas-Liquid Chromatography


“Peripheral” Properties of the Solvent Factors that usually don’t affect efficiency of the

separation but are of interest Synder defines a solvent “as either a pure

compound or a mixture of pure solvents.”

Binary and ternary solvent mixtures afford a wider range of solvents to choose than pure solvents

PERIPHERAL PROPERTIES

Boiling Point

• Properties• Easily evaporated or removed• bp 10-50 C higher than the temperature of

separation• Minimize accidental evaporation

– Diethyl ether

• Volatile samples – fractional distillation to remove solvent or sample

Boiling pt. (bp)Normally select a solvent with bp above that of the operation.Which distillation technique would you not want the above property?


ViscosityLow viscosity solvents preferable (General rule)

Liquid chromatography = poorer separationLow viscosity coincides with low bp

Exceptions: very polar solvents (alcohols) & compact molecules (cyclalkanes, aromatics, CCl4)

Low viscosity enhances diffusion which speed separation

Viscosity of a solvent mixture is usually intermediate between those of the pure solvents, ie binary mixture A & B

Take home – it is possible to use a viscous solvent when in a mixture

Peripheral Properties

Viscosity of Mixtures =(a)xa (b)xb


Viscosity of Water-Organic Solvent Mixtures


UV Cutoff -Solvent may interfere with detection

Appendix A shows minimum UV cut off for solvents

Solvent Properties Affecting Detection

What solvents might be poor choicesfor use with UV detection?


Refractive IndexMaximize differences in refractive index between sample and solvent (Appendix A)


Note the relatively small differencesWhat does this mean in relation to detection?


Specific Element ContentCommon Gas Chromatography Detectors:

Method Element SolventElectron Capture Cl ChloroformFlame Thermionic N AcetonitrileFlame Photometric S Dimethyl Sulfoxide


SOLVENT MISCIBILITY CHART

Peripheral Properties

Appendix C in your notes

– Toxicity– Flammability– Reactivity– Cost – Disposal


FACTORS AFFECTING SOLUBILITY AND SEPARATION

If the two solvents are immiscible, they can be shaken together and they will separate. If an analyte represented as ‘x’ is placed in one of the solvents before mixing, where will the analyte be after mixing?

Cx,a, Cx,b, concentration of solute x in solvents A & B;

R, gas constant (1.99 cal/oK);

T, temperature (oK);

G, free energy for transfer of 1 mole solute x from solvent B to A.

The concentration of x in the two solvents will be given as:

x

A

Bx

x

x x

x

x x

A

B

x

xx

x

x

x


H (enthalpy) change for transfer of 1 mole solute x from solvent B to A. If H positive, interaction with solvent B is stronger, the quantity on the right will be <1, and solute x will prefer phase B (Cx,b

> Cx,a)

In most solutions entropy effects are negligible: replace G with H

y = ex

0.001

0.01

0.1

1

10

100

-4 -2 0 2 4

Solvent Selection for Separation

Cx,b

Cx,a

= ?


Separation If solute x has a high solubility for the extracting

solvent while solute y has a low solubility, solute x will separate from solute y

The same principle applies to chromatography – solute x has high solubility for mobile phase while solute y has a high solubility for the stationary phase, therefore they will separate on the column – which one will move faster?


Solubility and Separation Visualize transfer of a

molecule x from solvent B to A, H = heat of transfer, and determines the relevant solvency of B vs. A for solute x.

Figure (a) portrays a part of molecule x

(i , functional group) with surrounding molecules of solvent B.


Figure (b) i is removed from solvent B leaving a cavity.

Figure (c ) The cavity collapses and B - i interactions are replaced with B - B interactions


Figure (d) original A- A interactions in Solvent A

Figure (e) the i group is added, breaking bonds between A molecules and forming a cavity

Figure (f) the i group is inserted into the cavity (dissolved)

WHAT GOVERNS THE STRENGTH OF THESE BONDS?

• Intermolecular Interactions– Dispersion Forces– Dipole-Dipole – Induced-Dipole– Hydrogen Bonding– Covalent Bonds

DISPERSION FORCESDispersion forces arise from the temporary variations in electron density around atoms and molecules. At any instant the electron distribution around an atom or molecule will likely produce a dipole moment, which can induce a (temporary) dipole moment in any nearby molecules. It is the Polarizability of the molecules, which determines the size of the induced dipole moments and thus the strength of the dispersion forces.

Molecules containing large atoms (e.g. bromine or iodine) have large polarizability and so give rise to large dispersion forces. This explains the increasing melting and boiling points of the halogens going down that group of the periodic table.

DISPERSION FORCES - SUMMARY

Polarizability- High Polarizability = High intermolecular attraction (larger atoms)

Molecular Size- Larger Size = More surface area and greater intermolecular attraction

Molecular Shape - More branching or compact shape has less surface area and lower intermolecular attraction.

DIPOLE - DIPOLE

If two neutral molecules, each having a permanent dipole moment, come together such that their oppositely charged ends align, they will be attracted to each other.

DIPOLE - DIPOLE

Orthodinitrobenzene has a high overall dipole moment because of the 2 nitro groups, but the overall dipole moment of the para compound is 0 because of the cancellation of group dipoles. However, both molecules have 2 nitro groups and the interactions of these two compounds with surrounding molecules are similar.

Interactions?

INDUCED DIPOLEA polar molecule (lower left) carries with it an electric field and this can induce a dipole moment in a nearby non-polar molecule (lower right). This will cause the attraction between the molecules.

This type of force is responsible for the solubility of oxygen (a non-polar molecule) in water (polar).

HYDROGEN BONDING

Hydrogen bonds are usually listed as a type of dipole-dipole force, but the details of hydrogen bonding are subtle and these bonds have some partial covalent bond character.

If a hydrogen bond can form between a pair of molecules it will be stronger than other intermolecular forces between the molecules.

Its time to play-Who wants to be

a millionaire?

FASTEST FINGER QUESTION

Place these molecules in orderfrom lowest to highest intermolecular forces

INTERMOLECULAR FORCES

Intermolecular Forces

Boiling Pt. oC1) neopentane 102) 2,3-dimethyl butane 583) n-hexane 694) 2-methyl-2-butanol 1025) 1-pentanol 138

POLARITY

Ability of a molecule to engage in strong interactions with other polar molecules. Thus, it describes the ability of the molecule to enter into many different interactions (dispersion, dipole, hydrogen bonding, etc.).

•Relative polarity – sum of all these interactions.

EXAMPLE:

Two Immiscible Solvents:

Solvent A WaterSolvent B Hexane

Analyte: Acetone (i)

How will the acetone partition between Solvent A and Solvent B?

EXAMPLE: USE THIS EQUATION

Cx,a

Cx,b

e -H/RT

What is the value of H?A

Bx

x

x x

x

x x

ESTIMATING THE VALUE OF H

2Hi,b = B – i bonds broken; --Hb,b = B – B bonds formed; Ha,a = A – A bonds broken; --2Hi,a = i – A bonds formed

H = (Ha,a – Hb,b) + 2(Hi,b – Hi,a)

H = (Pa2 - Pb

2) + 2Pi.(Pb – Pa)

Approximation:The interaction between moleculesis based on the product of their polarities. Thus:

CALCULATING H

H = (Pa2 - Pb

2) + 2Pi.(Pb – Pa)

All we need now are the polarities of A, B, and ito substitute in this equation:

From Appendix A:

i = Acetone P’ = 5.1 B = Hexane P’ = 0.1 A = Water P’ = 10.2

H = (10.22 - 0.12) + 10.2(0.1 – 10.2)

CALCULATING THE RATIO

Cx,a

Cx,b

e -H/RT

R, gas constant (1.99 cal/oK);

T, temperature (oK); = 298

H = ?

HYDROPHOBICITY

Hydrophobic interactions ‑ associated with "nonpolar" solutes in "polar" solvents

assume: B ‑ polar A ‑ non‑polar

i – small (non polar)

H = (Pa2 - Pb

2) + 2Pi.(Pb – Pa)

the polar solvent "squeezes" out the nonpolar solute into phase A.

SELECTIVITY

If there were only one type of interaction between molecules, the above equation for H would be valid. However, in reality there are usuallyseveral types of interactions.

H = (Pa2 - Pb

2) + 2Pi.(Pb – Pa)

These differences in interaction makes it possible to separate analytes of similar polarity.

This ability is known as solvent selectivity.

PART 2 SOLVENT CLASSIFICATION AND SELECTION

Outline:

Solvent Classification SchemesSummary of Solvent SelectionExtraction Efficiencies

HILDEBRAND SOLUBILITY PARAMETER

H = Vx [(a2 ‑ b

2) + 2x (b ‑ a)]

Vx (molar volume) of solute x affects its relative solubility

The larger is Vx more affected will be the solubility of x by a change in solvent polarity

ROHRSCHNEIDER POLARITY SCALE

The Rohrschneider polarity scale is based on experimental data. This method estimates the polarity of a solvent based on the solubility of three reference solutes below:

Ethanol proton donor interactionDioxane proton acceptor interactionNitromethane dipole interaction

ROHRSCHNEIDER POLARITY SCALE

The three values can be plotted as a Selectivity Triangle, with the 3 legs of the triangle calculated as the ratios of each individual term to the total polarity of the solvent as follows.

Xe = log(K"g) ethanol P’

Xd = log (K"g) dioxaneP’

Xn = log (K"g) nitromethane P’

SOLVENT SELECTIVITY TRIANGLE

EXTRACTION OF COMPOUND X FROM A SAMPLE MATRIX CONTAINING Y

Begin by studying the extraction of x and y as a function of solvent polarity.

EXTRACTION OF COMPOUND X FROM A SAMPLE MATRIX

x y

P’

% E

xtra

cted

P’1 P’x P’2 P’y

20

40

60

80

100

EXTRACTION OF COMPOUND Y FROM A SAMPLE MATRIX

x y

P’

% E

xtra

cted

P’1 P’x P’2 P’y

20

40

60

80

100

PARTITION COEFFICIENT

D or K = Co/Cw

Co is concentration in the organic phase (solvent)Cw is the concentration in the aqueous phase (water)

Simplest form of batch extraction

• Complete extraction not possible; greater than 99% extraction can occur

• Extraction efficiency by this method is based on Partition Coefficient (K) or Distribution ratio (D)

PARTITION COEFFICIENT

D = Co/Cw

Co is concentration in the organic phase (solvent)Cw is the concentration in the aqueous phase (water)

•Assume equal volumes

For unequal volumes, fraction extracted

DV1

DV

Vw Cw Vo Co

Vo Co

V = Vo/Vw

Fraction remaining in aqueous phase following n extractions: (1 - )n = Xn

If DV 100 then a single batch extraction can work:

Assume: V = Vo/Vw = 10D = Co/Cw = 5

Then:

= (5)(10)/1+(5)(10) = 98%

EXTRACTION EFFICIENCY

Wr = Wo (Vw/(KVo+ Vw))N

Where Wr = weight of solute remaining following extraction, Wo = weight of solute in original solution, Vw = volume of aqueous phase, Vo = volume of extracting solvent, K = partition coefficient, N = number of extractions.

Example: K = 2, Vw = 60 mL, Wo = 1 g Calculate Wr for 1 extraction with 60 mL solvent

2 extractions with 30 mL each 3 extractions with 20 mL each

Documents

S OLVENT S ELECTION FOR S EPARATION Processes Requiring Solvents Extraction Partition Fractional Crystallization Extractive Distillation Liquid Chromatography